A quadratic equation is a polynomial whose highest power is the square of a variable (x2, y2 etc.)

Definitions

A monomial is an algebraic expression with only one term in it.

Example: x3, 2x, y2, 3xyz etc.

A polynomial is an algebraic expression with more than one term in it.

Alternatively it can be stated as –

A polynomial is formed by adding/subtracting multiple monomials.

Example: x3+2y2+6x+10, 3x2+2x-1, 7y-2 etc.

A polynomial that contains two terms is called a binomial expression.

A polynomial that contains three terms is called a trinomial expression.

A standard quadratic equation looks like this:

ax2+bx+c = 0

Where a, b, c are numbers and a≥1.

a, b are called the coefficients of x2 and x respectively and c is called the constant.

The following are examples of some quadratic equations:

1) x2+5x+6 = 0 where a=1, b=5 and c=6.

2) x2+2x-3 = 0 where a=1, b=2 and c= -3

3) 3x2+2x = 1

→ 3x2+2x-1 = 0 where a=3, b=2 and c= -1

4) 9x2 = 4

→ 9x2-4 = 0 where a=9, b=0 and c= -4

For every quadratic equation, there can be one or more than one solution. These are called the roots of the quadratic equation.

For a quadratic equation ax2+bx+c = 0,

the sum of its roots = –b/a and the product of its roots = c/a.

A quadratic equation may be expressed as a product of two binomials.

For example, consider the following equation

x2-(a+b)x+ab = 0

x2-ax-bx+ab = 0

x(x-a)-b(x-a) = 0

(x-a)(x-b) = 0

x-a = 0 or x-b = 0
x = a or x=b

Here, a and b are called the roots of the given quadratic equation.

Now, let’s calculate the roots of an equation x2+5x+6 = 0.

We have to take two numbers adding which we get 5 and multiplying which we get 6. They are 2 and 3.

Let us express the middle term as an addition of 2x and 3x.

→ x2+2x+3x+6 = 0

→ x(x+2)+3(x+2) = 0

→ (x+2)(x+3) = 0

→ x+2 = 0 or x+3 = 0

→ x = -2 or x = -3

This method is called factoring.

We saw earlier that the sum of the roots is –b/a and the product of the roots is c/a. Let us verify that.

Sum of the roots for the equation x2+5x+6 = 0 is -5 and the product of the roots is 6.

The roots of this equation -2 and -3 when added give -5 and when multiplied give 6.

Solved examples of Quadratic equations

Let us solve some more examples using this method.

Problem 1: Solve for x: x2-3x-10 = 0

Solution:

Let us express -3x as a sum of -5x and +2x.

→ x2-5x+2x-10 = 0

→ x(x-5)+2(x-5) = 0

→ (x-5)(x+2) = 0

→ x-5 = 0 or x+2 = 0

→ x = 5 or x = -2

Problem 2: Solve for x: x2-18x+45 = 0

Solution:

The numbers which add up to -18 and give +45 when multiplied are -15 and -3.

Rewriting the equation,

→ x2-15x-3x+45 = 0

→ x(x-15)-3(x-15) = 0

→ (x-15) (x-3) = 0

→ x-15 = 0 or x-3 = 0

→ x = 15 or x = 3

Till now, the coefficient of x2 was 1. Let us see how to solve the equations where the coefficient of x2 is greater than 1.

Problem 3: Solve for x: 3x2+2x =1

Solution:

Rewriting our equation, we get 3x2+2x-1= 0

Here, the coefficient of x2 is 3. In these cases, we multiply the constant c with the coefficient of x2. Therefore, the product of the numbers we choose should be equal to -3 (-1*3).

Expressing 2x as a sum of +3x and –x

→ 3x2+3x-x-1 = 0

→ 3x(x+1)-1(x+1) = 0

→ (3x-1)(x+1) = 0

→ 3x-1 = 0 or x+1 = 0

→ x = 1/3 or x = -1

Problem 4: Solve for x: 11x2+18x+7 = 0

Solution:

In this case, the sum of the numbers we choose should equal to 18 and the product of the numbers should equal 11*7 = 77.

This can be done by expressing 18x as the sum of 11x and 7x.

→ 11x2+11x+7x+7 = 0

→ 11x(x+1) +7(x+1) = 0

→ (x+1)(11x+7) = 0

→ x+1 = 0 or 11x+7 = 0

→ x = -1 or x = -7/11.

The factoring method is an easy way of finding the roots. But this method can be applied only to equations that can be factored.

For example, consider the equation x2+2x-6=0.

If we take +3 and -2, multiplying them gives -6 but adding them doesn’t give +2. Hence this quadratic equation cannot be factored.

For this kind of equations, we apply the quadratic formula to find the roots.

The quadratic formula to find the roots,

x = [-b ± √(b2-4ac)] / 2a

Now, let us find the roots of the equation above.

x2+2x-6 = 0

Here, a = 1, b=2 and c= -6.

Substituting these values in the formula,

x = [-2 ± √(4 – (4*1*-6))] / 2*1

→ x = [-2 ± √(4+24)] / 2

→ x = [-2 ± √28] / 2

When we get a non-perfect square in a square root, we usually try to express it as a product of two numbers in which one is a perfect square. This is for simplification purpose. Here 28 can be expressed as a product of 4 and 7.

→ x = [-2 ± √(4*7)] / 2

→ x = [-2 ± 2√7] / 2

→ x = 2[ -1 ± √7] / 2

→ x = -1 ± √7

Hence, √7-1 and -√7-1 are the roots of this equation.

Let us consider another example.

Solve for x: x2 = 24 – 10x

Solution:

Rewriting the equation into the standard quadratic form,

x2 +10x-24 = 0

What are the two numbers which when added give +10 and when multiplied give -24? 12 and -2.

So this can be solved by the factoring method. But let’s solve it using the new method, applying the quadratic formula.

Here, a = 1, b = 10 and c = -24.

x = [-10 ± √(100 – 4*1*-24)] / 2*1

x = [-10 ± √(100-(-96))] / 2

x = [-10 ± √196] / 2

x = [-10 ± 14] / 2

x = 2 or x= -12 are the roots.

Discriminant

For an equation ax2+bx+c = 0, b2-4ac is called the discriminant and helps in determining the nature of the roots of a quadratic equation.

If b2-4ac > 0, the roots are real and distinct.

If b2-4ac = 0, the roots are real and equal.

If b2-4ac < 0, the roots are not real (they are complex).

Consider the following example:

Problem: Find the nature of roots for the equation x2+x+12 = 0.

Solution:

b2-4ac = -47 for this equation. So it has complex roots. Let us verify this.